Consecutive Integers

Question

If the first and third of three consecutive integers are added, the result is 57 less than five times the second integer. Find the third integer. 

Answer 1

First, re-read the problem until you understand it and can put it into your own words.  How's this?  There are three consecutive integers.  Five times the second integer minus 57 equals the sum of the first and third integer. What's the third number?
 
Then, if possible, assign a variable to the value the problem wants you to find (so you won't forget to solve for it later).  Here, let x = third integer.
 
Translate:
  "first and third ... are added, the result is 57 less than five times the second integer"
means     (x-2)  +  (x)                          =       - 57     +      5          *           (  x - 1 )
 
So, now the easy math:
     (x-2) + (x) = -57 + 5(x-1)                [rewrite equation]
           2x - 2   =  -57 + 5x - 5               [collect terms; distribute]
           -3x = -60                                  [subtract 5x from each side]
             x = 20                                 [divide each side by -3]
 
Checking:
  Is    18 + 20  =  -57 + 5*19    ?
           38  = -57 + 95   ?
            38  =  38   ? yes

Answer 2

For three consecutive integer, let.....

INT#1=x

INT#2=x+1

INT#3=x+2

 

Then.....

INT#1+INT#3=5(INT#2)-57...problem statement

 

Substitute above algebraic integer values into equation, get....

x+(x+2)=5(x+1)-57....solve for x...

2x+2=5x+5-57....simplify....

   -3x=-54

      x=18

 

Substitute into expressions for the three integers....

INT#1=x=18

INT#2=x+1=18+1=19

INT#3=x+2=18+2=20

 

The three consecutive integers are 18, 19 and 20.

 

Now check your work, substitute into problem parameters....

INT#1+INT#3=5(INT#2)-57

18+20=5(19)-57

      38=95-57

      38=38.....true, check√.  Your asumptions for consecutive integers

                     are correct and satisfy problem parameters

 

Answer 3

Let's assume the 3rd consecutive integer is unknown variable X, then we have the 2nd one (X-1) and the 1st one (X-2). Simple equation will help solving this math problem:

5(X-1)= X + (X-2) +57

 

Simplify

 

5X - 5 = 2X - 2 + 57

3X = 60

 

thus X=20